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Question: What does the equation \(\left( {a - b} \right)\left( {{x^2} + {y^2}} \right) - 2abx = 0\) become if...

What does the equation (a−b)(x2+y2)−2abx=0\left( {a - b} \right)\left( {{x^2} + {y^2}} \right) - 2abx = 0 become if the origin be moved to the point (aba−b,0)\left( {\dfrac{{ab}}{{a - b}},0} \right)?

Explanation

Solution

Let the origin be shifted to a point O'(h, k). If P(x, y) are coordinates of a point referred to old axes and P'(X, Y) are the coordinates of the same points referred to new axes, then x = X + h, y = Y + k.
We will apply the concept of 'Parallel Transformation of Coordinate Axes'. As the origin has been shifted to (aba−b,0)\left( {\dfrac{{ab}}{{a - b}},0} \right),so we can right x=X+aba−bx = X + \dfrac{{ab}}{{a - b}} and y=Yy = Y. Now we will substitute the value of x and y in the original equation and solve the problem.

Complete step by step solution:
The old equation of locus is (a−b)(x2+y2)−2abx=0\left( {a - b} \right)\left( {{x^2} + {y^2}} \right) - 2abx = 0.
Let a point on the curve with respect to origin be (x,y)\left( {x,y} \right)
Let the same point on the curve with respect to new origin be (X,Y)\left( {X,Y} \right)
The origin shifts from (0,0)\left( {0,0} \right)to (aba−b,0)\left( {\dfrac{{ab}}{{a - b}},0} \right).
Thus, from above statements we get
x=X+aba−bx = X + \dfrac{{ab}}{{a - b}}and y=Yy = Y
Now we will substitute the value of(x,y)\left( {x,y} \right)from above to the equation(a−b)(x2+y2)−2abx=0\left( {a - b} \right)\left( {{x^2} + {y^2}} \right) - 2abx = 0
⇒(a−b)((X+aba−b)2+Y2)−2ab(X+aba−b)=0\Rightarrow \left( {a - b} \right)\left( {{{\left( {X + \dfrac{{ab}}{{a - b}}} \right)}^2} + {Y^2}} \right) - 2ab\left( {X + \dfrac{{ab}}{{a - b}}} \right) = 0
Now we will open the brackets and expand the equation.
⇒(a−b)(X2+a2b2(a−b)2+2(Xaba−b)+Y2)−2abX−2a2b2a−b=0\Rightarrow \left( {a - b} \right)\left( {{X^2} + \dfrac{{{a^2}{b^2}}}{{{{\left( {a - b} \right)}^2}}} + 2\left( {\dfrac{{Xab}}{{a - b}}} \right) + {Y^2}} \right) - 2abX - \dfrac{{2{a^2}{b^2}}}{{a - b}} = 0
Now we will divided both sides by (a−b)\left( {a - b} \right)and we get:
⇒X2+a2b2(a−b)2+2Xaba−b+Y2−2Xaba−b−2a2b2(a−b)2=0\Rightarrow {X^2} + \dfrac{{{a^2}{b^2}}}{{{{\left( {a - b} \right)}^2}}} + 2\dfrac{{Xab}}{{a - b}} + {Y^2} - 2\dfrac{{Xab}}{{a - b}} - \dfrac{{2{a^2}{b^2}}}{{{{\left( {a - b} \right)}^2}}} = 0
Now cancelling the like terms 2Xaba−b2\dfrac{{Xab}}{{a - b}} and subtracting the like term a2b2(a−b)2\dfrac{{{a^2}{b^2}}}{{{{\left( {a - b} \right)}^2}}}, we get
⇒X2+Y2−a2b2(a−b)2=0\Rightarrow {X^2} + {Y^2} - \dfrac{{{a^2}{b^2}}}{{{{\left( {a - b} \right)}^2}}} = 0
Now we will add both sides by a2b2(a−b)2\dfrac{{{a^2}{b^2}}}{{{{\left( {a - b} \right)}^2}}}and we get:
⇒X2+Y2=a2b2(a−b)2\Rightarrow {X^2} + {Y^2} = \dfrac{{{a^2}{b^2}}}{{{{\left( {a - b} \right)}^2}}}
Now multiplying both sides by (a−b)2{\left( {a - b} \right)^2}, we get
⇒(X2+Y2)(a−b)2=a2b2\Rightarrow \left( {{X^2} + {Y^2}} \right){\left( {a - b} \right)^2} = {a^2}{b^2}

So, the equation (a−b)(x2+y2)−2abx=0\left( {a - b} \right)\left( {{x^2} + {y^2}} \right) - 2abx = 0 becomes (X2+Y2)(a−b)2=a2b2\left( {{X^2} + {Y^2}} \right){\left( {a - b} \right)^2} = {a^2}{b^2} if the origin be moved to the point (aba−b,0)\left( {\dfrac{{ab}}{{a - b}},0} \right).

Note:
The student can make an error if they don’t have prior knowledge of the method in which variables are replaced when the origin has been shifted. Another important aspect of this method is that new axes also remain parallel to the original ones, which is special about this process only.